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A fair coin is tossed 5 times. What is the probability

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A fair coin is tossed 5 times. What is the probability [#permalink]

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12 May 2007, 12:59
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A fair coin is tossed 5 times. What is the probability getting at least 3 heads on consecutive tosses?

a. 3/16
b. 1/4
c. 7/24
d. 5/16
e. 15/32
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12 May 2007, 13:22
It a hard problem, but I know how to solve it (I just learned binomial formula)

Ok the first step is to find combinations 5C3= 5!/(5-3)!*3!=5*4/2=10
Second is to find the probability of all wanted events, 1/2^3 (for all three it will be 1/8)
Third is to find the probability of all unwanted events, 1/2^2 (1/4)

now you have to multiply all these numbers

10/8*4

the answer is 10/32 or 5/16

P.S. sorry for a lame explanation, but it's very hard to explain (check the attached document)
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BASIC PROBABILITY.doc [184 KiB]

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12 May 2007, 20:17
Hi you answer would have been correct if the problem would have been "Three Heads" but I think you overlooked Consecutive heads...so the asnwer is 3/16.

Saurabh Malpani

Sergey_is_cool wrote:
It a hard problem, but I know how to solve it (I just learned binomial formula)

Ok the first step is to find combinations 5C3= 5!/(5-3)!*3!=5*4/2=10
Second is to find the probability of all wanted events, 1/2^3 (for all three it will be 1/8)
Third is to find the probability of all unwanted events, 1/2^2 (1/4)

now you have to multiply all these numbers

10/8*4

the answer is 10/32 or 5/16

P.S. sorry for a lame explanation, but it's very hard to explain (check the attached document)
Director
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12 May 2007, 21:15
saurabhmalpani wrote:
Hi you answer would have been correct if the problem would have been "Three Heads" but I think you overlooked Consecutive heads...so the asnwer is 3/16.

Saurabh Malpani

Sergey_is_cool wrote:
It a hard problem, but I know how to solve it (I just learned binomial formula)

Ok the first step is to find combinations 5C3= 5!/(5-3)!*3!=5*4/2=10
Second is to find the probability of all wanted events, 1/2^3 (for all three it will be 1/8)
Third is to find the probability of all unwanted events, 1/2^2 (1/4)

now you have to multiply all these numbers

10/8*4

the answer is 10/32 or 5/16

P.S. sorry for a lame explanation, but it's very hard to explain (check the attached document)

i got 1/8.

HHHHH
THHHH
TTHHH
TTHHH

I am not sure how you got 6?
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12 May 2007, 22:07
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The question stem asks not just about consecutive heads, but ''at least three'' consecutive heads: this means, of course, the sum of the probobility of 3 consecutive heads, the probability of 4 consecutive heads, and the probability of 5 consecutive heads. So--Himalayan-- you have these possibilies: HHHTT, THHHT, TTHHH, HHHHT, THHHH, and HHHHH. This ends up resulting in a probability of 6/32 = 3/16.
Director
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13 May 2007, 04:03
clue - count all cases where we can get the three or more consecutive heads.
Director
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13 May 2007, 12:50

Probability of atleast 3 = prob 3 + prob 4 + prob 5

Probability of 3 consecutive heads = 3 x (1/2)^5

(1st,2nd,3rd or 2nd,3rd,4th or 3rd,4th,5th)

Probability of 4 cons. heads = 2 x (1/2)^5

(1st,2nd,3rd,4th or 2nd,3rd,4th,5th)

Probability of 5 cons. heads = 1 x (1/2)^5

Total probability = 6 x (1/2)^5 = 3/16

Is there any direct formula to calculate the above ?

say probability of atleast 49 consecutive heads in 200 tosses .. ?
Director
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13 May 2007, 15:44
OA is 'B'.

Total ways = 2^5 =32
Ways of getting at least 3 consecutive heads are:8
HHHTT
HHHHT
HHHHH
TTHHH
THHHT
THHHH
HTHHH
HHHTH

hence probability = 8/32 =1/4
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13 May 2007, 15:49
Oh Shit missed two cases!!!

Thanks for pointing out the mistake.

Saurabh Malpani

vshaunak@gmail.com wrote:
OA is 'B'.

Total ways = 2^5 =32
Ways of getting at least 3 consecutive heads are:8
HHHTT
HHHHT
HHHHH
TTHHH
THHHT
THHHH
HTHHH
HHHTH

hence probability = 8/32 =1/4
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